三角関数の還元(負角・余角・補角)公式
三角関数の還元公式
負角
余角
補角
負角
(1)
\[ \sin(-x)=-\sin x \](2)
\[ \cos(-x)=\cos x \](3)
\[ \tan(-x)=-\tan x \]余角
(4)
\[ \sin\left(\frac{\pi}{2}-x\right)=\cos x \](5)
\[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \](6)
\[ \tan\left(\frac{\pi}{2}-x\right)=\cot x \]補角
(7)
\[ \sin\left(\pi-x\right)=\sin x \](8)
\[ \cos\left(\pi-x\right)=-\cos x \](9)
\[ \tan\left(\pi-x\right)=-\tan x \](1)
\begin{align*} \sin(-x) & =\frac{e^{-ix}-e^{ix}}{2i}\\ & =-\frac{e^{ix}-e^{-ix}}{2i}\\ & =-\sin x \end{align*}(2)
\begin{align*} \cos(-x) & =\frac{e^{-ix}+e^{ix}}{2}\\ & =\cos x \end{align*}(3)
\begin{align*} \tan(-x) & =\frac{\sin(-x)}{\cos(-x)}\\ & =\frac{-\sin(x)}{\cos x}\\ & =-\tan x \end{align*}(4)
\begin{align*} \sin\left(\frac{\pi}{2}-x\right) & =\sin\frac{\pi}{2}\cos x-\cos\frac{\pi}{2}\sin x\\ & =\cos x \end{align*}(5)
\begin{align*} \cos\left(\frac{\pi}{2}-x\right) & =\cos\frac{\pi}{2}\cos x+\sin\frac{\pi}{2}\sin x\\ & =\sin x \end{align*}(6)
\begin{align*} \tan\left(\frac{\pi}{2}-x\right) & =\frac{\sin\left(\frac{\pi}{2}-x\right)}{\cos\left(\frac{\pi}{2}-x\right)}\\ & =\frac{\cos x}{\sin x}\\ & =\cot x \end{align*}(7)
\begin{align*} \sin\left(\pi-x\right) & =\sin\pi\cos x-\cos\pi\sin x\\ & =\sin x \end{align*}(8)
\begin{align*} \cos\left(\pi-x\right) & =\cos\pi\cos x+\sin\pi\sin x\\ & =-\cos x \end{align*}(9)
\begin{align*} \tan\left(\pi-x\right) & =\frac{\sin(\pi-x)}{\cos(\pi-x)}\\ & =\frac{\sin x}{-\cos x}\\ & =-\tan x \end{align*}ページ情報
タイトル | 三角関数の還元(負角・余角・補角)公式 |
URL | https://www.nomuramath.com/rj4agkrd/ |
SNSボタン |
x tan(x)とx tanh(x)の積分
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\int z\tan^{\pm1}\left(z\right)dz=i^{\pm1}\left\{ \frac{1}{2}z^{2}-iz\Li_{1}\left(\mp e^{2iz}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{2iz}\right)\right\} +C
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\[
\Sin^{\bullet}\left(iz\right)=i\Sinh^{\bullet}z
\]
三角関数と双曲線関数の冪乗積分漸化式
\[
\int\sin^{n}xdx=-\frac{1}{n}\cos x\sin^{n-1}x+\frac{n-1}{n}\int\sin^{n-2}xdx\qquad(n\ne0)
\]
三角関数と双曲線関数の積和公式と和積公式
\[ \sin\alpha\cos\beta=\frac{1}{2}\left\{ \sin(\alpha+\beta)+\sin(\alpha-\beta)\right\}
\]